---
title: "Shaft torque measurement using strain gauges"
subject: "Instrumentation and Measurement"
chapter: 7
key: "semester_4/instrumentation_and_measurement:shaft-torque-strain-gauge-numerical"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/shaft-torque-strain-gauge-numerical
site: BCT Engineering
---

# Shaft torque measurement using strain gauges

Instrumentation and Measurement · Chapter 7: Common Mechanical Measurement System and Transducers

- Asked in: year not given, 8 marks
- Video: [Torque measurement using strain guage](https://www.youtube.com/watch?v=TD5XbHmIXU8) — e Learning Together
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/shaft-torque-strain-gauge-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-7

## Question

Explain how the torque on a rotating shaft is measured using strain gauges. A 40 mm diameter solid steel shaft (G = 80 GPa) carries four gauges at $\pm 45^\circ$ to the axis, connected as a full bridge. Each gauge shows a strain of magnitude 400 microstrain. Find the shear stress, torque and the power transmitted at 1500 rpm. If the gauge factor is 2.1 and the excitation is 10 V, find the bridge output.

## Short answer

Torque creates pure shear, so gauges at ±45 degrees measure principal strains; a full bridge cancels bending/axial effects and adds shear strain. Shear stress=2Gε=64 MPa, torque=804 N m, power at 1500 rpm=126.3 kW, and bridge output=8.4 mV.

## Answer

#### Torque measurement by strain gauges
A shaft in torsion is in pure shear at its surface. The principal stresses are $\pm\tau$ along directions at $45^\circ$ to the axis (tension along one diagonal, compression along the other). Four gauges are bonded at $\pm45^\circ$, two on each diagonal, and connected as a full bridge so that all four outputs add. This arrangement is insensitive to axial load and bending (their strains cancel in the bridge) and compensates for temperature. The bridge supply and output are carried to and from the rotating shaft by slip rings, or by a telemetry or rotary-transformer system.

```text
   shaft axis ------------------------>
        /  \      gauges at +45 and -45 degrees
       / R1 \     R1, R3 : +45 (tension)
      / R2   \    R2, R4 : -45 (compression)
```

Strain at $45^\circ$ in pure shear:

$$
\varepsilon_{45} = \frac{\gamma}{2} = \frac{\tau}{2G}
$$

The torque is then found from $\tau = 16T/(\pi d^3)$.

#### Calculation
Given: $d = 40$ mm, $G = 80$ GPa, $\varepsilon = 400\ \mu\varepsilon$, $N = 1500$ rpm.

Shear stress:

$$
\tau = 2G\varepsilon = 2(80\times10^{9})(400\times10^{-6}) = 64\times10^{6}\ \text{Pa} = 64\ \text{MPa}
$$

Torque:

$$
T = \frac{\pi d^3\tau}{16} = \frac{\pi(0.04)^3(64\times10^{6})}{16} = 804.2\ \text{N m}
$$

Power:

$$
P = \frac{2\pi N T}{60} = \frac{2\pi(1500)(804.2)}{60} = 126.3\times10^{3}\ \text{W} = 126.3\ \text{kW}
$$

Bridge output (all four arms active):

$$
V_o = V_{ex}\,G_f\,\varepsilon = 10 \times 2.1 \times 400\times10^{-6} = 8.4\ \text{mV}
$$

**Answer: $\tau = 64$ MPa, $T = 804$ N m, $P = 126.3$ kW, bridge output $= 8.4$ mV.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why are the gauges placed at ±45° to the shaft axis?
   A. Principal stresses in pure shear act at 45°, tension on one diagonal and compression on the other
   B. Axial stress is maximum there
   C. Bending is maximum there
   D. It is only convenient
   Answer: A. Torsion gives pure shear with principal stresses ±tau at 45° to the axis, giving the largest gauge strains. (Key point: Gauge orientation)

2. With G = 80 GPa and 45° strain of 400 microstrain, shear stress is (e45 = tau/2G):
   A. 32 MPa
   B. 64 MPa
   C. 128 MPa
   D. 16 MPa
   Answer: B. tau = 2 G e = 2 x 80e9 x 400e-6 = 64 MPa. (Key point: Shear stress)

3. For d = 40 mm and tau = 64 MPa, torque T = pi d^3 tau / 16 is about:
   A. 402 N m
   B. 804 N m
   C. 1608 N m
   D. 80.4 N m
   Answer: B. T = pi x (0.04)^3 x 64e6 / 16 = 804.2 N m. (Key point: Torque)

4. With all four arms active, G = 2.1, 10 V and 400 microstrain, bridge output is:
   A. 2.1 mV
   B. 4.2 mV
   C. 8.4 mV
   D. 84 mV
   Answer: C. Vo = Vex G e = 10 x 2.1 x 400e-6 = 8.4 mV. (Key point: Bridge output)

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